4 network design cfvg 2012
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11SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Modeling Supply Chain
&
Network Planning
Prof. Ravi Shankar
Department of Management Studies,
Indian Institute of Technology Delhi,
New Delhi
Prof. Ravi Shankar
Department of Management Studies,
Indian Institute of Technology Delhi,
New Delhi
“Supply Chain Management”
44SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Example#1: Transportation, Problems
�Can be formulated as linear programs and solved by general purpose linear programming codes.
� However, there are many computer packages, which contain separate computer codes for these models which take advantage of their network structure.
55SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Transportation Problem
�The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si ) to n destinations (each with a demand dj ), when the unit shipping cost from an origin, i, to a destination, j, is cij.
66SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
N-DELHI
S-DELHI
E-DELHI
W-DELHI
Destinations
Sources
FARIDABAD
GURGAON
GAHZIABAD
S1=250
S2=200
S3= 250
D1=100
D2=150
D3=250
D4=100
17
19
16
17
13
19
16
15
1517
17
16
CASE STUDY#1
77SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
TP: a Linear Programming Model
•The structure of the model is:
� Minimize Total Shipping Cost� ST
� [Amount shipped from a source] <= [Supply at that source]
� [Amount received at a destination]=[Demand at that Destination]
• Decision variables� Xij = the number of cases shipped from plant i to warehouse j.
� where: i=1 (FARIDABAD), 2 (GURGAON), 3 (GHAZIABAD)
j=1 (N-DELHI), 2 (S-DELHI), 3 (E-DELHI), 4(W-DELHI)
88SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
N-Delhi
S-Delhi
E-Delhi
W-Delhi
D1=100
D2=150
D3=250
D4=100
The supply constraints
Faridabad
S1=250
X11
X12
X13
X14
Supply from FARIDABAD X11+X12+X13+X14 = 250
Gurgaon
S2=200
X21
X22
X23
X24
Supply from GURGAON X21+X22+X23+X24 = 200
GhaziabadS3= 250
X31
X32
X33
X34
Supply from Ghaziabad X31+X32+X33+X34 = 250
CASE STUDY
99SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
The complete mathematical model
Minimize 13X11+16X12+19X13+ 17X14 +17X21+19X22+16X23+15X24+
15X31+17X32+17X33+16X34
ST
Supply constrraints:
X11+ X12+ X13+ X14 250
X21+ X22+ X23+ X24 200
X31+ X32+ X33+ X34 250
Demand constraints:
X11+ X21+ X31 100
X12+ X22+ X32 150
X13+ X23+ X33 250
X14+ X24+ X34 100
All Xij are nonnegative
≤
≤
≤
=
=
=
=
Total shipment out of a supply nodecannot exceed the supply at the node.
Total shipment received at a destinationnode, must equal the demand at that node.
CASE STUDY
1010SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
The complete mathematical model
Minimize 13X11+16X12+19X13+ 17X14 +17X21+19X22+16X23+15X24+
15X31+17X32+17X33+16X34
ST
Supply constrraints:
X11+ X12+ X13+ X14 250
X21+ X22+ X23+ X24 200
X31+ X32+ X33+ X34 250
Demand constraints:
X11+ X21+ X31 100
X12+ X22+ X32 150
X13+ X23+ X33 250
X14+ X24+ X34 100
All Xij are nonnegative
≤
≤
≤
=
=
=
=
Total shipment out of a supply nodecannot exceed the supply at the node.
Total shipment received at a destinationnode, must equal the demand at that node.
CASE STUDY#6
1313SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
– Reduced costs • The unit shipment cost between
Gurgaon and N-Delhi must be reduced by at least Rs.5, before it would become economically feasible to utilize it
• If this route is used, the total cost will increase by Rs. 5 for each case shipped between Gurgaon and N-Delhi .
1515SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
– Allowable Increase/Decrease• This is the range of optimality.
• The unit shipment cost between Gurgaon and N-Delhi may increase up to any level or decrease up to Rs. 5 with no change in the current optimal transportation plan.
2020SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Solution Techniques
� Mathematical optimization techniques:
1. Exact algorithms: find optimal solutions
2. Heuristics: find “good” solutions, not necessarily optimal
� Simulation models: provide a mechanism to evaluate specified design alternatives created by the designer.
2121SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Example
� Single product
� Two plants p1 and p2
• Plant p2 has an annual capacity of 60,000 units.
� The two plants have the same production costs.
� There are two warehouses w1 and w2 with identical warehouse handling costs.
� There are three markets areas c1,c2 and c3 with demands of 50,000, 100,000 and 50,000, respectively.
2222SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Unit Distribution Costs
Facility
warehouse
p1 p2 c1 c2 c3
w1 0 4 3 4 5
w2 5 2 2 1 2
2323SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
CASE : NETWORK PROBLEM
� Network Representation
ARNOLD
WASH
BURN
ZROX
HEWES
200,000200,000
60,000060,0000
50,00050,000
100,000100,000
50,00050,000
00
55
44
22
33
44
55
22 11
22
P1P1
P2P2
C2C2
C1C1
W1W1
W2W2
C3C3
2424SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Heuristic #1:Choose the Cheapest Warehouse to Source Demand
D = 50,000
D = 100,000
D = 50,000
Cap = 60,000
$5 x 140,000
$2 x 60,000
$2 x 50,000
$1 x 100,000
$2 x 50,000
Total Costs = $1,120,000
2525SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Heuristic #2:Choose the warehouse where the total delivery costs to and from the warehouse are the lowest[Consider inbound and outbound distribution costs]
D = 50,000
D = 100,000
D = 50,000
Cap = 60,000
$4
$5
$2
$3
$4$5
$2
$1
$2
$0
P1 to WH1 $3P1 to WH2 $7P2 to WH1 $7P2 to WH 2 $4
P1 to WH1 $4P1 to WH2 $6P2 to WH1 $8P2 to WH 2 $3
P1 to WH1 $5P1 to WH2 $7P2 to WH1 $9P2 to WH 2 $4
Market #1 is served by WH1, Markets 2 and 3are served by WH2
2626SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
D = 50,000
D = 100,000
D = 50,000
Cap = 60,000
Cap = 200,000
$5 x 90,000
$2 x 60,000
$3 x 50,000
$1 x 100,000
$2 x 50,000
$0 x 50,000
P1 to WH1 $3P1 to WH2 $7P2 to WH1 $7P2 to WH 2 $4
P1 to WH1 $4P1 to WH2 $6P2 to WH1 $8P2 to WH 2 $3
P1 to WH1 $5P1 to WH2 $7P2 to WH1 $9P2 to WH 2 $4
Total Cost = $920,000
Heuristic #2:Choose the warehouse where the total delivery costs to and from the warehouse are the lowest[Consider inbound and outbound distribution costs]
2727SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
The Optimization Model
The problem described earlier can be framed as the following linear programming problem.
Let
� x(p1,w1), x(p1,w2), x(p2,w1) and x(p2,w2) be the flows from the plants to the warehouses.
� x(w1,c1), x(w1,c2), x(w1,c3) be the flows from the warehouse w1 to customer zones c1, c2 and c3.
� x(w2,c1), x(w2,c2), x(w2,c3) be the flows from warehouse w2 to customer zones c1, c2 and c3
2828SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
The problem we want to solve is: min 0x(p1,w1) + 5x(p1,w2) + 4x(p2,w1)
+ 2x(p2,w2) + 3x(w1,c1) + 4x(w1,c2)
+ 5x(w1,c3) + 2x(w2,c1) + 2x(w2,c3)
subject to the following constraints:x(p2,w1) + x(p2,w2) ≤ 60000
x(p1,w1) + x(p2,w1) = x(w1,c1) + x(w1,c2) + x(w1,c3)
x(p1,w2) + x(p2,w2) = x(w2,c1) + x(w2,c2) + x(w2,c3)
x(w1,c1) + x(w2,c1) = 50000
x(w1,c2) + x(w2,c2) = 100000
x(w1,c3) + x(w2,c3) = 50000
all flows greater than or equal to zero.
The Optimization Model
2929SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
CASE : NETWORK DESIGN
� Network Representation
ARNOLD
WASH
BURN
ZROX
HEWES
200,000200,000
60,00060,000
50,00050,000
100,000100,000
50,00050,000
0 (A)0 (A)
5 (B)5 (B)
4 (C)4 (C)
2 (D) 2 (D)
3 (E)3 (E)
4 (F)4 (F)
5 (G)5 (G)
2 (H)2 (H)1 (I)1 (I)
2 (J)2 (J)
P1P1
P2P2
C2C2
C1C1
W1W1
W2W2
C3C3
3030SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
The problem we want to solve is: min 0x(p1,w1) + 5x(p1,w2) + 4x(p2,w1)
+ 2x(p2,w2) + 3x(w1,c1) + 4x(w1,c2)
+ 5x(w1,c3) + 2x(w2,c1) + 2x(w2,c3)
subject to the following constraints:x(p2,w1) + x(p2,w2) ≤ 60000
x(p1,w1) + x(p2,w1) = x(w1,c1) + x(w1,c2) + x(w1,c3)
x(p1,w2) + x(p2,w2) = x(w2,c1) + x(w2,c2) + x(w2,c3)
x(w1,c1) + x(w2,c1) = 50000
x(w1,c2) + x(w2,c2) = 100000
x(w1,c3) + x(w2,c3) = 50000
all flows greater than or equal to zero.
The Optimization Model
3131SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Optimal Solution
Facility
warehouse
p1 p2 c1 c2 c3
w1 140,000 0 50,000 40,000 50,000
w2 0 60,000 0 60,000 0
Total cost for the optimal strategy is $740,000
3232SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Optimal Solution
Facility
warehouse
p1 p2 c1 c2 c3
w1 140,000 0 50,000 40,000 50,000
w2 0 60,000 0 60,000 0
Total cost for the optimal strategy is $740,000
3333SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
CASE : NETWORK PROBLEM
� Network Representation
ARNOLD
WASH
BURN
ZROX
HEWES
200,000200,000
60,00060,000
50,00050,000
100,000100,000
50,00050,000
00
55
44
22
33
44
55
22 11
22
P1P1
P2P2
C2C2
C1C1
W1W1
W2W2
C3C3
140,000140,000
60,00060,000
50,00050,000
40,,00040,,000
50,00050,000
60,000
60,000
3434SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
New Supply Chain Strategy
� OBJECTIVES:• Reduce inventory and financial risks• Provide customers with competitive response times.
� ACHIEVE THE FOLLOWING:• Determining the optimal location of inventory across the various
stages • Calculating the optimal quantity of safety stock for each component at
each stage
� Hybrid strategy of Push and Pull• Push Stages produce to stock where the company keeps safety stock• Pull stages keep no stock at all.
� Challenge:• Identify the location where the strategy switched from Push-based to
Pull-based• Identify the Push-Pull boundary
� Benefits:• For same lead times, safety stock reduced by 40 to 60%• Company could cut lead times to customers by 50% and still reduce
safety stocks by 30%
3535SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Three Different Product Categories
� High variability - low volume products
� Low variability - high volume products, and
� Low variability - low volume products.
3636SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Supply Chain Strategy Different for the Different Categories
� High variability low volume products • Position them mainly at the primary warehouses
� demand from many retail outlets can be aggregated reducing inventory costs.
� Low variability high volume products • Position close to the retail outlets at the secondary warehouses
• Ship fully loaded tracks as close as possible to the customers reducing transportation costs.
� Low variability low volume products • Require more analysis since other characteristics are important, such as profit margins, etc.